\displaystyle \textbf{Question 1: }\text{The sum of two numbers is }60 \text{ and their difference is }14.
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }x\text{ and }y.
\displaystyle x+y=60\qquad\text{...(i)}
\displaystyle x-y=14\qquad\text{...(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle (x+y)+(x-y)=60+14
\displaystyle \Rightarrow 2x=74
\displaystyle \Rightarrow x=37
\displaystyle \text{Substituting }x=37\text{ in equation (i),}
\displaystyle 37+y=60
\displaystyle \Rightarrow y=23
\displaystyle \therefore \text{ the two numbers are }37\text{ and }23.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Twice a number is equal to thrice}
\displaystyle \text{the other number. If the sum of the numbers is }85, \text{ find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }x\text{ and }y.
\displaystyle 2x=3y\qquad\text{...(i)}
\displaystyle x+y=85\qquad\text{...(ii)}
\displaystyle \text{Rewriting equation (i),}
\displaystyle 2x-3y=0\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 2x+2y=170\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (2x-3y)-(2x+2y)=0-170
\displaystyle \Rightarrow -5y=-170
\displaystyle \Rightarrow y=34
\displaystyle \text{Substituting }y=34\text{ in equation (i),}
\displaystyle 2x=3(34)
\displaystyle \Rightarrow x=\frac{102}{2}=51
\displaystyle \therefore \text{ the two numbers are }51\text{ and }34.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find two numbers such that twice the}
\displaystyle \text{first added to thrice the second gives }70 \text{ and twice the second}
\displaystyle \text{added to thrice the first gives }75.
\displaystyle \text{Answer:}
\displaystyle \text{Let the first number be }x\text{ and the second number be }y.
\displaystyle 2x+3y=70\qquad\text{...(i)}
\displaystyle 3x+2y=75\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 6x+9y=210\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 6x+4y=150\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (6x+9y)-(6x+4y)=210-150
\displaystyle \Rightarrow 5y=60
\displaystyle \Rightarrow y=12
\displaystyle \text{Substituting }y=12\text{ in equation (i),}
\displaystyle 2x+3(12)=70
\displaystyle \Rightarrow 2x=34
\displaystyle \Rightarrow x=17
\displaystyle \therefore \text{ the two numbers are }17\text{ and }12.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find two numbers which differ by }9 \text{ and are such that}
\displaystyle \text{four times the larger added to three times the smaller gives }92.
\displaystyle \text{Answer:}
\displaystyle \text{Let the larger number be }x\text{ and the smaller number be }y.
\displaystyle x-y=9\qquad\text{...(i)}
\displaystyle 4x+3y=92\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }4,\text{ we get}
\displaystyle 4x-4y=36\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle (4x-4y)-(4x+3y)=36-92
\displaystyle \Rightarrow -7y=-56
\displaystyle \Rightarrow y=8
\displaystyle \text{Substituting }y=8\text{ in equation (i),}
\displaystyle x-8=9
\displaystyle \Rightarrow x=17
\displaystyle \therefore \text{ the two numbers are }17\text{ and }8.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The sum of two numbers is }30 \text{ and the difference of their}
\displaystyle \text{squares is }180.\text{ Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }x\text{ and }y.
\displaystyle x+y=30\qquad\text{...(i)}
\displaystyle x^2-y^2=180\qquad\text{...(ii)}
\displaystyle \text{Using }x^2-y^2=(x-y)(x+y)\text{ in equation (ii),}
\displaystyle (x-y)(x+y)=180
\displaystyle \text{Using }x+y=30,\text{ we get}
\displaystyle 30(x-y)=180
\displaystyle \Rightarrow x-y=6\qquad\text{...(iii)}
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle (x+y)+(x-y)=30+6
\displaystyle \Rightarrow 2x=36
\displaystyle \Rightarrow x=18
\displaystyle \text{Substituting }x=18\text{ in equation (i),}
\displaystyle 18+y=30
\displaystyle \Rightarrow y=12
\displaystyle \therefore \text{ the two numbers are }18\text{ and }12.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The sum of the digits of a two-digit number is }8.
\displaystyle \text{On adding }18\text{ to the number, its digits are reversed.}
\displaystyle \text{Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the tens digit and }y\text{ be the units digit.}
\displaystyle \text{Therefore, the two-digit number is }10x+y.
\displaystyle x+y=8\qquad\text{...(i)}
\displaystyle \text{On reversing the digits, the number becomes }10y+x.
\displaystyle (10x+y)+18=10y+x\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 10x+y+18=10y+x
\displaystyle \Rightarrow 9x-9y=-18
\displaystyle \Rightarrow x-y=-2\qquad\text{...(iii)}
\displaystyle \text{Adding equations (i) and (iii),}
\displaystyle (x+y)+(x-y)=8+(-2)
\displaystyle \Rightarrow 2x=6
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (i),}
\displaystyle 3+y=8
\displaystyle \Rightarrow y=5
\displaystyle \therefore \text{ the required number is }10(3)+5=35.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A two-digit number is three times the sum}
\displaystyle \text{of its digits. If }45\text{ is added to the number, its digits are}
\displaystyle \text{reversed. Find the original number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the tens digit and }y\text{ be the units digit.}
\displaystyle \text{Therefore, the two-digit number is }10x+y.
\displaystyle 10x+y=3(x+y)\qquad\text{...(i)}
\displaystyle 10x+y+45=10y+x\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (i),}
\displaystyle 10x+y=3x+3y
\displaystyle \Rightarrow 7x-2y=0\qquad\text{...(iii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 9x-9y=-45
\displaystyle \Rightarrow x-y=-5\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iv) by }7,\text{ we get}
\displaystyle 7x-7y=-35\qquad\text{...(v)}
\displaystyle \text{Subtracting equation (v) from equation (iii),}
\displaystyle (7x-2y)-(7x-7y)=0-(-35)
\displaystyle \Rightarrow 5y=35
\displaystyle \Rightarrow y=7
\displaystyle \text{Substituting }y=7\text{ in equation (iv),}
\displaystyle x-7=-5
\displaystyle \Rightarrow x=2
\displaystyle \therefore \text{ the original number is }10(2)+7=27.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{A two-digit number is seven times the sum}
\displaystyle \text{of its digits. If }27\text{ is subtracted from the number, its digits}
\displaystyle \text{get interchanged. Find the number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ be the tens digit and }y\text{ be the units digit.}
\displaystyle \text{Therefore, the two-digit number is }10x+y.
\displaystyle 10x+y=7(x+y)\qquad\text{...(i)}
\displaystyle 10x+y-27=10y+x\qquad\text{...(ii)}
\displaystyle \text{Simplifying equation (i),}
\displaystyle 10x+y=7x+7y
\displaystyle \Rightarrow 3x-6y=0\qquad\text{...(iii)}
\displaystyle \text{Simplifying equation (ii),}
\displaystyle 9x-9y=27
\displaystyle \Rightarrow x-y=3\qquad\text{...(iv)}
\displaystyle \text{Multiplying equation (iv) by }3,\text{ we get}
\displaystyle 3x-3y=9\qquad\text{...(v)}
\displaystyle \text{Subtracting equation (v) from equation (iii),}
\displaystyle (3x-6y)-(3x-3y)=0-9
\displaystyle \Rightarrow -3y=-9
\displaystyle \Rightarrow y=3
\displaystyle \text{Substituting }y=3\text{ in equation (iv),}
\displaystyle x-3=3
\displaystyle \Rightarrow x=6
\displaystyle \therefore \text{ the required number is }10(6)+3=63.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find a fraction which reduces to }\frac{2}{3} \text{ when }3\text{ is added}
\displaystyle \text{to both its numerator and denominator, and reduces to }\frac{3}{5} \text{ when }1
\displaystyle \text{is added to both its numerator and denominator.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required fraction be }\frac{x}{y}.
\displaystyle \frac{x+3}{y+3}=\frac{2}{3}
\displaystyle \Rightarrow 3(x+3)=2(y+3)
\displaystyle \Rightarrow 3x+9=2y+6
\displaystyle \Rightarrow 3x-2y=-3\qquad\text{...(i)}
\displaystyle \frac{x+1}{y+1}=\frac{3}{5}
\displaystyle \Rightarrow 5(x+1)=3(y+1)
\displaystyle \Rightarrow 5x+5=3y+3
\displaystyle \Rightarrow 5x-3y=-2\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }5,\text{ we get}
\displaystyle 15x-10y=-15\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,\text{ we get}
\displaystyle 15x-9y=-6\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (15x-10y)-(15x-9y)=-15-(-6)
\displaystyle \Rightarrow -y=-9
\displaystyle \Rightarrow y=9
\displaystyle \text{Substituting }y=9\text{ in equation (i),}
\displaystyle 3x-2(9)=-3
\displaystyle \Rightarrow 3x=15
\displaystyle \Rightarrow x=5
\displaystyle \therefore \text{ the required fraction is }\frac{5}{9}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{On adding }1\text{ to the numerator of a}
\displaystyle \text{fraction, it becomes }\frac{1}{2}. \text{ Also, on adding }1\text{ to the denominator}
\displaystyle \text{of the original fraction, it becomes }\frac{1}{3}. \text{ Find the original fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the original fraction be }\frac{x}{y}.
\displaystyle \frac{x+1}{y}=\frac{1}{2}
\displaystyle \Rightarrow 2(x+1)=y
\displaystyle \Rightarrow 2x-y=-2\qquad\text{...(i)}
\displaystyle \frac{x}{y+1}=\frac{1}{3}
\displaystyle \Rightarrow 3x=y+1
\displaystyle \Rightarrow 3x-y=1\qquad\text{...(ii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle (3x-y)-(2x-y)=1-(-2)
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (ii),}
\displaystyle 3(3)-y=1
\displaystyle \Rightarrow 9-y=1
\displaystyle \Rightarrow y=8
\displaystyle \therefore \text{ the original fraction is }\frac{3}{8}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In a given fraction, if the numerator is}
\displaystyle \text{multiplied by }2\text{ and the denominator is reduced by }5, \text{ we get }\frac{6}{5}.
\displaystyle \text{But, if the numerator is increased by }8 \text{ and the denominator is}
\displaystyle \text{doubled, we get }\frac{2}{5}. \text{ Find the fraction.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required fraction be }\frac{x}{y}.
\displaystyle \frac{2x}{y-5}=\frac{6}{5}
\displaystyle \Rightarrow 10x=6(y-5)
\displaystyle \Rightarrow 10x-6y=-30\qquad\text{...(i)}
\displaystyle \frac{x+8}{2y}=\frac{2}{5}
\displaystyle \Rightarrow 5(x+8)=4y
\displaystyle \Rightarrow 5x-4y=-40\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 10x-8y=-80\qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (i),}
\displaystyle (10x-6y)-(10x-8y)=-30-(-80)
\displaystyle \Rightarrow 2y=50
\displaystyle \Rightarrow y=25
\displaystyle \text{Substituting }y=25\text{ in equation (ii),}
\displaystyle 5x-4(25)=-40
\displaystyle \Rightarrow 5x-100=-40
\displaystyle \Rightarrow 5x=60
\displaystyle \Rightarrow x=12
\displaystyle \therefore \text{ the required fraction is }\frac{12}{25}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Five years ago, a lady was thrice as old}
\displaystyle \text{as her daughter. Ten years hence, the lady would be twice as old}
\displaystyle \text{as her daughter. What are their present ages?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of the lady be }x\text{ years.}
\displaystyle \text{Let the present age of her daughter be }y\text{ years.}
\displaystyle x-5=3(y-5)
\displaystyle \Rightarrow x-5=3y-15
\displaystyle \Rightarrow x-3y=-10\qquad\text{...(i)}
\displaystyle x+10=2(y+10)
\displaystyle \Rightarrow x+10=2y+20
\displaystyle \Rightarrow x-2y=10\qquad\text{...(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle (x-3y)-(x-2y)=-10-10
\displaystyle \Rightarrow -y=-20
\displaystyle \Rightarrow y=20
\displaystyle \text{Substituting }y=20\text{ in equation (ii),}
\displaystyle x-2(20)=10
\displaystyle \Rightarrow x=50
\displaystyle \therefore \text{ the lady's present age is }50\text{ years,}
\displaystyle \text{and her daughter's present age is }20\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The sum of the ages of A and B is }39 \text{ years. In }15\text{ years'}
\displaystyle \text{time, the age of A will be twice the age of B. Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of A be }x\text{ years.}
\displaystyle \text{Let the present age of B be }y\text{ years.}
\displaystyle x+y=39\qquad\text{...(i)}
\displaystyle x+15=2(y+15)
\displaystyle \Rightarrow x+15=2y+30
\displaystyle \Rightarrow x-2y=15\qquad\text{...(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle (x+y)-(x-2y)=39-15
\displaystyle \Rightarrow 3y=24
\displaystyle \Rightarrow y=8
\displaystyle \text{Substituting }y=8\text{ in equation (i),}
\displaystyle x+8=39
\displaystyle \Rightarrow x=31
\displaystyle \therefore \text{ A's present age is }31\text{ years,}
\displaystyle \text{and B's present age is }8\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{A is }15\text{ years older than B. Five years ago,}
\displaystyle \text{A was four times as old as B. Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of B be }x\text{ years.}
\displaystyle \therefore \text{ the present age of A is }(x+15)\text{ years.}
\displaystyle \text{Five years ago, B's age was }(x-5)\text{ years,}
\displaystyle \text{and A's age was }(x+15-5)=(x+10)\text{ years.}
\displaystyle x+10=4(x-5)
\displaystyle \Rightarrow x+10=4x-20
\displaystyle \Rightarrow 3x=30
\displaystyle \Rightarrow x=10
\displaystyle \therefore \text{ B's present age is }10\text{ years.}
\displaystyle \text{A's present age}=10+15=25\text{ years.}
\displaystyle \therefore \text{ A's present age is }25\text{ years,}
\displaystyle \text{and B's present age is }10\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Six years ago, the ages of Geeta and Seema}
\displaystyle \text{were in the ratio }3:4.\text{ Nine years hence, their ages will be}
\displaystyle \text{in the ratio }6:7.\text{ Find their present ages.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the present age of Geeta be }x\text{ years.}
\displaystyle \text{Let the present age of Seema be }y\text{ years.}
\displaystyle \frac{x-6}{y-6}=\frac{3}{4}
\displaystyle \Rightarrow 4(x-6)=3(y-6)
\displaystyle \Rightarrow 4x-24=3y-18
\displaystyle \Rightarrow 4x-3y=6\qquad\text{...(i)}
\displaystyle \frac{x+9}{y+9}=\frac{6}{7}
\displaystyle \Rightarrow 7(x+9)=6(y+9)
\displaystyle \Rightarrow 7x+63=6y+54
\displaystyle \Rightarrow 7x-6y=-9\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }7,\text{ we get}
\displaystyle 28x-21y=42\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }4,\text{ we get}
\displaystyle 28x-24y=-36\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (28x-21y)-(28x-24y)=42-(-36)
\displaystyle \Rightarrow 3y=78
\displaystyle \Rightarrow y=26
\displaystyle \text{Substituting }y=26\text{ in equation (i),}
\displaystyle 4x-3(26)=6
\displaystyle \Rightarrow 4x-78=6
\displaystyle \Rightarrow 4x=84
\displaystyle \Rightarrow x=21
\displaystyle \therefore \text{ Geeta's present age is }21\text{ years,}
\displaystyle \text{and Seema's present age is }26\text{ years.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Four knives and six forks cost Rs. }200, \text{ while six knives}
\displaystyle \text{and seven forks together cost Rs. }264. \text{ Find the cost of a knife and a fork.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost of a knife be Rs. }x.
\displaystyle \text{Let the cost of a fork be Rs. }y.
\displaystyle 4x+6y=200\qquad\text{...(i)}
\displaystyle 6x+7y=264\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }3,\text{ we get}
\displaystyle 12x+18y=600\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle 12x+14y=528\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (12x+18y)-(12x+14y)=600-528
\displaystyle \Rightarrow 4y=72
\displaystyle \Rightarrow y=18
\displaystyle \text{Substituting }y=18\text{ in equation (i),}
\displaystyle 4x+6(18)=200
\displaystyle \Rightarrow 4x=92
\displaystyle \Rightarrow x=23
\displaystyle \therefore \text{ the cost of a knife is Rs. }23,
\displaystyle \text{and the cost of a fork is Rs. }18.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The cost of }13\text{ cups and }16 \text{ spoons is Rs. }296,\text{ while the cost}
\displaystyle \text{of }16\text{ cups and }13\text{ spoons is Rs. }284. \text{ Find the cost of }2\text{ cups and }5\text{ spoons.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the cost of a cup be Rs. }x.
\displaystyle \text{Let the cost of a spoon be Rs. }y.
\displaystyle 13x+16y=296\qquad\text{...(i)}
\displaystyle 16x+13y=284\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }16,\text{ we get}
\displaystyle 208x+256y=4736\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }13,\text{ we get}
\displaystyle 208x+169y=3692\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (208x+256y)-(208x+169y)=4736-3692
\displaystyle \Rightarrow 87y=1044
\displaystyle \Rightarrow y=12
\displaystyle \text{Substituting }y=12\text{ in equation (i),}
\displaystyle 13x+16(12)=296
\displaystyle \Rightarrow 13x=104
\displaystyle \Rightarrow x=8
\displaystyle \text{Therefore, the cost of }2\text{ cups and }5\text{ spoons is}
\displaystyle 2x+5y=2(8)+5(12)
\displaystyle =16+60=76.
\displaystyle \therefore \text{ the required cost is Rs. }76.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Rahul covered a distance of }128\text{ km in }5 \text{ hours, partly on a bicycle}
\displaystyle \text{at }16\text{ km/h and partly on a moped at }32\text{ km/h. How much distance did he cover}
\displaystyle \text{on the moped?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the distance covered on the bicycle be }x\text{ km.}
\displaystyle \therefore \text{ the distance covered on the moped is }(128-x)\text{ km.}
\displaystyle \text{Using }\text{Time}=\frac{\text{Distance}}{\text{Speed}},
\displaystyle \frac{x}{16}+\frac{128-x}{32}=5
\displaystyle \text{Multiplying throughout by }32,\text{ we get}
\displaystyle 2x+128-x=160
\displaystyle \Rightarrow x+128=160
\displaystyle \Rightarrow x=32
\displaystyle \therefore \text{ distance covered on the bicycle}=32\text{ km.}
\displaystyle \text{Distance covered on the moped}=128-32=96\text{ km.}
\displaystyle \therefore \text{ Rahul covered }96\text{ km on the moped.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{A boat can go }75\text{ km downstream in }5 \text{ hours and }44\text{ km}
\displaystyle \text{upstream in }4\text{ hours. Find (i) the speed of the boat in still water}
\displaystyle \text{and (ii) the rate of the current.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of the boat in still water be }x\text{ km/h.}
\displaystyle \text{Let the speed of the current be }y\text{ km/h.}
\displaystyle \text{Downstream speed}=(x+y)\text{ km/h.}
\displaystyle \frac{75}{x+y}=5
\displaystyle \Rightarrow 75=5(x+y)
\displaystyle \Rightarrow x+y=15\qquad\text{...(i)}
\displaystyle \text{Upstream speed}=(x-y)\text{ km/h.}
\displaystyle \frac{44}{x-y}=4
\displaystyle \Rightarrow 44=4(x-y)
\displaystyle \Rightarrow x-y=11\qquad\text{...(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle (x+y)+(x-y)=15+11
\displaystyle \Rightarrow 2x=26
\displaystyle \Rightarrow x=13
\displaystyle \text{Substituting }x=13\text{ in equation (i),}
\displaystyle 13+y=15
\displaystyle \Rightarrow y=2
\displaystyle \therefore \text{ (i) Speed of the boat in still water}=13\text{ km/h.}
\displaystyle \text{(ii) Rate of the current}=2\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The monthly incomes of A and B are in the ratio }4:3 \text{ and their monthly}
\displaystyle \text{savings are in the ratio }9:5. \text{ If each spends Rs. }3500\text{ per month, find the}
\displaystyle \text{monthly income of each.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the monthly income of A be Rs. }x.
\displaystyle \text{Let the monthly income of B be Rs. }y.
\displaystyle \frac{x}{y}=\frac{4}{3}
\displaystyle \Rightarrow 3x=4y\qquad\text{...(i)}
\displaystyle \text{A's monthly saving is Rs. }(x-3500).
\displaystyle \text{B's monthly saving is Rs. }(y-3500).
\displaystyle \frac{x-3500}{y-3500}=\frac{9}{5}
\displaystyle \Rightarrow 5(x-3500)=9(y-3500)
\displaystyle \Rightarrow 5x-17500=9y-31500
\displaystyle \Rightarrow 5x-9y=-14000\qquad\text{...(ii)}
\displaystyle \text{From equation (i), }x=\frac{4}{3}y.
\displaystyle \text{Substituting }x=\frac{4}{3}y\text{ in equation (ii),}
\displaystyle 5\left(\frac{4}{3}y\right)-9y=-14000
\displaystyle \Rightarrow \frac{20y}{3}-9y=-14000
\displaystyle \Rightarrow \frac{20y-27y}{3}=-14000
\displaystyle \Rightarrow -7y=-42000
\displaystyle \Rightarrow y=6000
\displaystyle \therefore x=\frac{4}{3}(6000)=8000
\displaystyle \therefore \text{ A's monthly income is Rs. }8000,
\displaystyle \text{and B's monthly income is Rs. }6000.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Six nuts and five bolts weigh }278\text{ grams, while eight nuts and three}
\displaystyle \text{bolts weigh }268\text{ grams. Find the weight of each nut and each bolt.}
\displaystyle \text{How much do three nuts and three bolts weigh together?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the weight of each nut be }x\text{ g.}
\displaystyle \text{Let the weight of each bolt be }y\text{ g.}
\displaystyle 6x+5y=278\qquad\text{...(i)}
\displaystyle 8x+3y=268\qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (i) by }8,\text{ we get}
\displaystyle 48x+40y=2224\qquad\text{...(iii)}
\displaystyle \text{Multiplying equation (ii) by }6,\text{ we get}
\displaystyle 48x+18y=1608\qquad\text{...(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle (48x+40y)-(48x+18y)=2224-1608
\displaystyle \Rightarrow 22y=616
\displaystyle \Rightarrow y=28
\displaystyle \text{Substituting }y=28\text{ in equation (i),}
\displaystyle 6x+5(28)=278
\displaystyle \Rightarrow 6x=138
\displaystyle \Rightarrow x=23
\displaystyle \therefore \text{ the weight of each nut is }23\text{ g,}
\displaystyle \text{and the weight of each bolt is }28\text{ g.}
\displaystyle \text{Weight of three nuts and three bolts}
\displaystyle =3(23)+3(28)
\displaystyle =69+84=153\text{ g}.
\displaystyle \therefore \text{ three nuts and three bolts weigh }153\text{ g.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{There are some girls in two classrooms, A and B.}
\displaystyle \text{If }12\text{ girls are sent from room A to room B, the number of girls}
\displaystyle \text{in both rooms will become equal. If }11\text{ girls are sent from room B}
\displaystyle \text{to room A, then the number of girls in room A would be double the}
\displaystyle \text{number of girls in room B. How many girls are there in each classroom?}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of girls in classroom A be }x.
\displaystyle \text{Let the number of girls in classroom B be }y.
\displaystyle x-12=y+12
\displaystyle \Rightarrow x-y=24\qquad\text{...(i)}
\displaystyle x+11=2(y-11)
\displaystyle \Rightarrow x+11=2y-22
\displaystyle \Rightarrow x-2y=-33\qquad\text{...(ii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle (x-y)-(x-2y)=24-(-33)
\displaystyle \Rightarrow y=57
\displaystyle \text{Substituting }y=57\text{ in equation (i),}
\displaystyle x-57=24
\displaystyle \Rightarrow x=81
\displaystyle \therefore \text{ the number of girls in classroom A is }81,
\displaystyle \text{and the number of girls in classroom B is }57.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Four men and four boys can do a piece of work}
\displaystyle \text{in }3\text{ days, while }2\text{ men and }5\text{ boys can finish it in }4 \text{ days.}
\displaystyle \text{How long would it take one man alone to do it?}
\displaystyle \text{Answer:}
\displaystyle \text{Suppose one man alone can finish the work in }x\text{ days,}
\displaystyle \text{and one boy alone can finish the work in }y\text{ days.}
\displaystyle \therefore \text{ one man's one day's work}=\frac{1}{x}.
\displaystyle \text{One boy's one day's work}=\frac{1}{y}.
\displaystyle \text{Four men and four boys can finish the work in }3\text{ days.}
\displaystyle \therefore \frac{4}{x}+\frac{4}{y}=\frac{1}{3} \qquad\text{...(i)}
\displaystyle \text{Two men and five boys can finish the work in }4\text{ days.}
\displaystyle \therefore \frac{2}{x}+\frac{5}{y}=\frac{1}{4} \qquad\text{...(ii)}
\displaystyle \text{Multiplying equation (ii) by }2,\text{ we get}
\displaystyle \frac{4}{x}+\frac{10}{y}=\frac{1}{2} \qquad\text{...(iii)}
\displaystyle \text{Subtracting equation (i) from equation (iii),}
\displaystyle \frac{6}{y}=\frac{1}{2}-\frac{1}{3}
\displaystyle \Rightarrow \frac{6}{y}=\frac{1}{6}
\displaystyle \Rightarrow y=36
\displaystyle \text{Substituting }y=36\text{ in equation (i),}
\displaystyle \frac{4}{x}+\frac{4}{36}=\frac{1}{3}
\displaystyle \Rightarrow \frac{4}{x}=\frac{1}{3}-\frac{1}{9}
\displaystyle \Rightarrow \frac{4}{x}=\frac{2}{9}
\displaystyle \Rightarrow 2x=36
\displaystyle \Rightarrow x=18
\displaystyle \therefore \text{ one man alone can finish the work in }18\text{ days.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{A takes }3\text{ hours longer than B to walk }30 \text{ km. But, if A doubles}
\displaystyle \text{his pace, he is ahead of B by }1\text{ hour }30\text{ minutes. Find the speeds of A and B.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the speed of A be }x\text{ km/h.}
\displaystyle \text{Let the speed of B be }y\text{ km/h.}
\displaystyle \text{Time taken by A to walk }30\text{ km}=\frac{30}{x}\text{ hours.}
\displaystyle \text{Time taken by B to walk }30\text{ km}=\frac{30}{y}\text{ hours.}
\displaystyle \frac{30}{x}=\frac{30}{y}+3\qquad\text{...(i)}
\displaystyle \text{If A doubles his speed, his speed becomes }2x\text{ km/h.}
\displaystyle \therefore \text{ his time taken becomes }\frac{30}{2x} =\frac{15}{x}\text{ hours.}
\displaystyle 1\text{ hour }30\text{ minutes}=\frac{3}{2}\text{ hours.}
\displaystyle \frac{15}{x}+\frac{3}{2}=\frac{30}{y} \qquad\text{...(ii)}
\displaystyle \text{From equation (i), }\frac{30}{y} =\frac{30}{x}-3.
\displaystyle \text{Substituting this in equation (ii),}
\displaystyle \frac{15}{x}+\frac{3}{2}=\frac{30}{x}-3
\displaystyle \Rightarrow \frac{30}{x}-\frac{15}{x} =3+\frac{3}{2}
\displaystyle \Rightarrow \frac{15}{x}=\frac{9}{2}
\displaystyle \Rightarrow 9x=30
\displaystyle \Rightarrow x=\frac{10}{3}=3\frac{1}{3}\text{ km/h}.
\displaystyle \text{Substituting }x=\frac{10}{3}\text{ in equation (i),}
\displaystyle \frac{30}{10/3}=\frac{30}{y}+3
\displaystyle \Rightarrow 9=\frac{30}{y}+3
\displaystyle \Rightarrow \frac{30}{y}=6
\displaystyle \Rightarrow y=5
\displaystyle \therefore \text{ the speed of A is }3\frac{1}{3}\text{ km/h,}
\displaystyle \text{and the speed of B is }5\text{ km/h.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If the length of a rectangle is reduced by }1 \text{ m and its breadth is}
\displaystyle \text{increased by }2\text{ m, its area increases by }32\text{ m}^2. \text{ If, however, the length is}
\displaystyle \text{increased by }2\text{ m and its breadth is reduced by }3\text{ m, then its area is reduced}
\displaystyle \text{by }49\text{ m}^2. \text{ Find the length and breadth of the rectangle.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the rectangle be }x\text{ m.}
\displaystyle \text{Let the breadth of the rectangle be }y\text{ m.}
\displaystyle \therefore \text{ the original area}=xy\text{ m}^2.
\displaystyle (x-1)(y+2)=xy+32
\displaystyle \Rightarrow xy+2x-y-2=xy+32
\displaystyle \Rightarrow 2x-y=34\qquad\text{...(i)}
\displaystyle (x+2)(y-3)=xy-49
\displaystyle \Rightarrow xy-3x+2y-6=xy-49
\displaystyle \Rightarrow -3x+2y=-43
\displaystyle \Rightarrow 3x-2y=43\qquad\text{...(ii)}
\displaystyle \text{From equation (i),}
\displaystyle y=2x-34
\displaystyle \text{Substituting }y=2x-34\text{ in equation (ii),}
\displaystyle 3x-2(2x-34)=43
\displaystyle \Rightarrow 3x-4x+68=43
\displaystyle \Rightarrow -x=-25
\displaystyle \Rightarrow x=25
\displaystyle \text{Substituting }x=25\text{ in equation (i),}
\displaystyle y=2(25)-34=16
\displaystyle \therefore \text{ the length of the rectangle is }25\text{ m,}
\displaystyle \text{and the breadth of the rectangle is }16\text{ m.}
\displaystyle \\

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